Question: Perform the row operation, $-6R_3\rightarrow R_3$, on the following matrix. $\left[\begin{array} {ccc} -4 & 6 & 4 & 0 \\ 2 & -7 & -8 & 0 \\ 2 & -3 & 1 &- 7 \end{array} \right] $
Background There are three basic row operations that can be performed on matrices. $R_i \leftrightarrow R_j$. This symbol tells us to interchange rows $i$ and $j$. $cR_i \rightarrow R_i$. This symbol tells us to multiply a row $i$ by a constant $c$. $R_i + cR_j \rightarrow R_i$. This symbol tells us to add $c$ times row $j$ to row $i$. Finding the new row to be used For the given matrix, $R_3$ is given below. $R_3=\left[\begin{array} {ccc} 2 & -3 & 1 &- 7\end{array} \right]$ We are asked to perform the row operation, $-6R_3\rightarrow R_3$. Therefore, we must multiply $R_3$ by $-6$. $\begin{aligned}-6R_3 &= -6\left[\begin{array} {ccc} 2 & -3 & 1 &- 7 \end{array} \right] \\\\&=\left[\begin{array} {ccc} -12 & 18 & -6 & 42 \end{array} \right]\end{aligned}$ Substituting the row Now, we must substitute row $R_3$ with $-6R_3$. $\left[\begin{array} {ccc} -4 & 6 & 4 & 0 \\ 2 & -7 & -8 & 0 \\ {2} & {-3} & {1} & {-7} \end{array} \right]\xrightarrow{-6R_3\rightarrow R_3} \left[\begin{array} {ccc} -4 & 6 & 4 & 0 \\ 2 & -7 & -8 & 0 \\ {-12} & {18} & {-6} & {42} \end{array} \right]$ Summary Our resultant matrix is the following. $\left[\begin{array} {ccc} -4 & 6 & 4 & 0 \\ 2 & -7 & -8 & 0 \\ -12 & 18 & -6 & 42 \end{array} \right]$